NEET 2025 — Physics solution

NEET2025Physics

The kinetic energies of two similar cars A and B are $100$ J and $225$ J respectively. On applying brakes, car A stops after $1000$ m and car B stops after $1500$ m. If $F_{A}$ and $F_{B}$ are the forces applied by the brakes on cars A and B respectively, then the ratio $F_{A} / F_{B}$ is:

A
$\frac{3}{2}$
B
$\frac{2}{3}$
C
$\frac{1}{3}$
D
$\frac{1}{2}$

Solution

By the work-energy theorem, the work done by the braking force equals the change in kinetic energy: $F \cdot S = Δ K E$ .

For each car, $- F \cdot S = 0 - K E_{i}$ , i.e. $F S = K E_{i}$ .

$\frac{F _{A}}{F _{B}} = \frac{K E _{A} / S _{A}}{K E _{B} / S _{B}} = \frac{100/1000}{225/1500} = \frac{0.1}{0.15} = \frac{2}{3}$ .

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